# Finding the Inverse. 1 st example, begin with your function f(x) = 3x – 7 replace f(x) with y y = 3x - 7 Interchange x and y to find the inverse x = 3y.

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Finding the Inverse

1 st example, begin with your function f(x) = 3x – 7 replace f(x) with y y = 3x - 7 Interchange x and y to find the inverse x = 3y – 7 now solve for y x + 7 = 3y = y f -1 (x) = replace y with f -1 (x)

2 nd example g(x) = 2x 3 + 1 replace g(x) with y y = 2x 3 + 1 Interchange x and y to find the inverse x = 2y 3 + 1 now solve for y x - 1 = 2y 3 = y 3 = y g -1 (x) = replace y with g -1 (x)

Recall, to verify you have found the inverse you check that composition of the function with the inverse, in both orders, equals x Using specific ordered pairs can illustrate how the inverse works, but does not verify that it is the inverse.

Consider f(x) = What is the domain? x + 4 > 0 x > -4 or the interval [-4, ∞) What is the range? y > 0 or the interval [0, ∞)

Now find the inverse: f(x) =D: [-4, ∞) R: [0, ∞) y = Interchange x and y x = x 2 = y + 4 x 2 – 4 = y f -1 (x) = x 2 – 4D: [0, ∞) R: [-4, ∞)

Finally, let us consider the graphs: f(x) = D: [-4, ∞) R: [0, ∞) blue graph f -1 (x) = x 2 – 4 D: [0, ∞) R: [-4, ∞) red graph

2 nd example Consider g(x) = 5 - x 2 D: [0, ∞) What is the range? Make a very quick sketch of the graph R: (-∞, 5]

Now find the inverse: g(x) = 5 - x 2 D: [0, ∞) R: (-∞, 5] y = 5 - x 2 Interchange x and y x = 5 - y 2 x – 5 = -y 2 5 – x = y 2 = y but do we want the + or – square root? g -1 (x) = D: (-∞, 5] R: [0, ∞)

And, now the graphs: g(x) = 5 - x 2 D: [0, ∞) R: (-∞, 5] blue graph g -1 (x) = D: (-∞, 5] R: [0, ∞) red graph

A function is one-to-one if each x and y-value is unique Algebraically it means if f(a)=f(b), then a=b. On a graph it means the graph passes the vertical and the horizontal line tests. If a function is one-to-one it has an inverse function.

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